Is there any result about the calculation of radius of tubular neighborhood of submanifold inside a Riemannian manifold?

For example, given a simple smooth curve on R^2, what's the radius of its tubular neighborhood? (One upper bound is given by the minimal curvature, but general it is not the radius)

Maybe that is what we can expect:
if the curvature of the curve is always decreasing, then the radius of the tubular neighborhood is given by the injective radius of (left) end point, it seems true, right?

I do not see what answer one can expect; given the obstruction drawn below, of which you seem to be aware, in terms of what quantities would you like the radius of the tubular neighborhood be expressed?
–
Benoît KloecknerOct 26 '12 at 12:36

2 Answers
2

As has already been pointed out, while the tubular radius bounds the curvature globally from below, the curvature information is not enough to correctly estimate this radius. Many examples can be considered, often silly ones when you allow your sub-manifold to be disconnected: consider the case of two parallel line segments in the plane and note that the embedding radius depends on their distance.

Since it is not clear what exactly you are after (computing the tubular radius of the curve $xy = 1$ is not so hard) or what information you already have, I would like to mention that your problem is solvable by calculus alone! Whether this calculus is tractable or not obviously depends on your choice of Riemannian manifold $X$ and submanifold $M \subset X$.

Given $M \subset X$, the Medial Axis $A_X(M)$ of $M$ in $X$ is defined to be the collection of all $x \in X \setminus M$ such that there are multiple solutions to the following constrained optimization problem in $X$:

Minimize $\text{dist}_X(x,m)$ subject to $m \in M$.

Here is a rough sketch of what a medial axis looks like when $X = \mathbb{R}^2$ and $M$ is the Nicolaescu horseshoe. The axis itself is in blue, and the red lines are my amateurish attempts at showing equidistant $M$-points

Once you know this medial axis, the distance from $A_X(M)$ to $M$ is precisely your tubular radius. Figuring out this distance again reduces to calculus which may be intractably hard depending on the choice of $X$ and $M$.

Update Here is a simple pictorial counterexample to the claim that if the curvature is decreasing from "left to right" then the tubular radius is the injectivity radius of the "left" endpoint. A straight line going up a lampshade suffices. If you want the curvature to decrease strictly, you can wrap the initial segment of the curve around the lower edge of the lampshade and then straighten as you go up. The point is that the curvature of the ambient manifold also plays a part in restricting the tubular radius. It is likely that your conjecture applies in Euclidean space, although I don't have an immediate proof of this.

The inverse of the curvature of a plane curve $C$ at a point $p$ is the radius of the osculator circle to the curve at the point $p$; see e.g. Geometry and Imagination by Hilbert and Cohn-Vossen. This suggests that the largest radius of a tube ought to be the inverse of the maximum of the curvature. This is optimal for example when the curve is a circle. There are however possible global obstructions, when two points, far apart along the curve, are actually really close in $\mathbb{R}^2$; think a horseshoe.

actually this is exactly what the OP is saying...
–
Pietro MajerOct 26 '12 at 12:33

Thanks a lot! Maybe that is what I want: if the curvature of the curve is always decreasing, then the radius of the tubular neighborhood is given by the injective radius of (left) end point, it seems true, right?
–
JayOct 26 '12 at 15:13

If it decreases going one way along the curve, it increases going the other way. The example of a spiral given in polar coordinates by the equation $r=e^{-\theta}$, $0\leq \theta \leq 2\pi N$, may provide a counterexample to your conjectured formula. It seems plausible that if the curve is the $C^2$-boundary of a convex domain then the largest tubular radius is the inverse of the maximal curvature.
–
Liviu NicolaescuOct 26 '12 at 15:33

Let me write it in a different way $x(t)=e^{-t}\cos t$ $y(t)=e^{-t}\sin t$, $t\in [0,2\pi N]$. As $t$ runs in the interval $[0,2\pi N]$ the point $(x(t),y(t))$ keeps on orbiting around the origin, while the distance to the origin decays exponentially to $0$. But I know believe it may not be a counterexample to your conjecture.
–
Liviu NicolaescuOct 26 '12 at 17:39